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932. Beautiful Array

Description

An array nums of length n is beautiful if:

  • nums is a permutation of the integers in the range [1, n].
  • For every 0 <= i < j < n, there is no index k with i < k < j where 2 * nums[k] == nums[i] + nums[j].

Given the integer n, return any beautiful array nums of length n. There will be at least one valid answer for the given n.

 

Example 1:

Input: n = 4
Output: [2,1,4,3]

Example 2:

Input: n = 5
Output: [3,1,2,5,4]

 

Constraints:

  • 1 <= n <= 1000

Solutions

Solution 1

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class Solution:
    def beautifulArray(self, n: int) -> List[int]:
        if n == 1:
            return [1]
        left = self.beautifulArray((n + 1) >> 1)
        right = self.beautifulArray(n >> 1)
        left = [x * 2 - 1 for x in left]
        right = [x * 2 for x in right]
        return left + right

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class Solution {
    public int[] beautifulArray(int n) {
        if (n == 1) {
            return new int[] {1};
        }
        int[] left = beautifulArray((n + 1) >> 1);
        int[] right = beautifulArray(n >> 1);
        int[] ans = new int[n];
        int i = 0;
        for (int x : left) {
            ans[i++] = x * 2 - 1;
        }
        for (int x : right) {
            ans[i++] = x * 2;
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> beautifulArray(int n) {
        if (n == 1) return {1};
        vector<int> left = beautifulArray((n + 1) >> 1);
        vector<int> right = beautifulArray(n >> 1);
        vector<int> ans(n);
        int i = 0;
        for (int& x : left) ans[i++] = x * 2 - 1;
        for (int& x : right) ans[i++] = x * 2;
        return ans;
    }
};

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func beautifulArray(n int) []int {
    if n == 1 {
        return []int{1}
    }
    left := beautifulArray((n + 1) >> 1)
    right := beautifulArray(n >> 1)
    var ans []int
    for _, x := range left {
        ans = append(ans, x*2-1)
    }
    for _, x := range right {
        ans = append(ans, x*2)
    }
    return ans
}

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